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Consider the following functions $F_{ij}:S\subset{\mathbb R}^3\to{\mathbb R}$, $$ F_{ij}(y) = \begin{cases} \frac{(y_i-x_i)(y_j-x_j)(y-x)\cdot n(y)}{|y-x|^3},&y\neq x; \\ 0,& y=x.\end{cases} \quad i,j = 1,2,3 $$ where $S$ is a surface which has a continuously varying normal vector, $x=(x_1,x_2,x_3)\in S$ is given, $y=(y_1,y_2,y_3)\in S$, $n(y)$ is the normal vector at point $y$ [EDITED: which is assumed to be smooth]. Here $(y-x)\cdot n(y)$ is the dot product. Using the method in the answer to the question on MO, I conclude that given $i,j$ $$ \lim_{y\to x}F_{ij}(y)=0 $$ which implies that $F_{ij}(y)$ is continuous at $y=x$.

Here is my question:

  • Is $F_{ij}(y)$ smooth at $x$? If it is not, what would be the key properties to fail the smoothness?

An immediate idea is that I should test the smoothness of $F_{ij}$ by definition. The difficulty is that with a parameterization $y=y(\alpha,\beta)$, it is not trivial to find the high order partial derivatives for $F_{ij}(y(\alpha,\beta))$. I am not even able to determine if $F_{ij}$ is $C^1(S)$.


[EDITED:] Using the parameterization in this answer, I am able to get something like $$ F_{12}(\alpha,\beta)=\frac{-\alpha\beta (\alpha h_{\alpha}+\beta h_{\beta}-h)}{(\alpha^2+\beta^2+h^2)^{3/2}} $$ where $y(\alpha,\beta)=(\alpha,\beta,h(\alpha,\beta))$. Due to the root term, I guess $F_{12}$ can not be $C^1$.

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You need that $n(x)$ is smooth in this case. –  azarel Mar 19 '12 at 19:30
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